Consider the Dirac quantum field $$\psi(x) = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{\sqrt{2E_{\mathbb{p}}}} \sum_{s} [a^{s}_{\mathbb{p}}u^{s}(\mathbb{p})e^{-\mathrm{i}p \cdot x} + b^{s\dagger}_{\mathbb{p}}v^{s}(\mathbb{p})e^{+\mathrm{i}p \cdot x}],$$ using the conventions of Peskin and Schroeder. Due to the anticommutation relations $$\{a^{r}_{\mathbb{p}},a^{s\dagger}_{\mathbb{q}} \} = (2\pi)^{3}\delta^{(3)}(\mathbb{p}-\mathbb{q})\delta^{rs}, \\ \{b^{r\dagger}_{\mathbb{p}},b^{s\dagger}_{\mathbb{q}} \} = (2\pi)^{3}\delta^{(3)}(\mathbb{p}-\mathbb{q})\delta^{rs},$$ all other anticommutators being zero; and the spin sums $$\sum_{s} u^{s}(\mathbb{p})u^{s\dagger}(\mathbb{p}) = (\gamma^{\rho} p_{\rho} + m) \gamma^{0}, \\ \sum_{s} v^{s}(\mathbb{p})v^{s\dagger}(\mathbb{p}) = (\gamma^{\rho} p_{\rho} - m) \gamma^{0},$$ it obeys $$\{ \psi_{a}(x),\psi^{\dagger}_{b}(y) \} = [(\mathrm{i}\gamma^{\rho}\partial_{\rho\vert x} + m)\gamma^{0}]_{ab} \Delta(x-y),$$ where $\partial_{\rho\vert x}$ means differentiation with respect to $x$, and where $$\Delta(x-y) = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E_{\mathbb{p}}}[e^{-\mathrm{i}p \cdot (x-y)} - e^{+\mathrm{i}p \cdot (x-y)}].$$ This implies causality: $\{ \psi_{a}(x),\psi^{\dagger}_{b}(y) \} = 0$, for $x-y$ space-like, because $\Delta(x-y) = 0$ then.
Consider next the Lorentz transformation $\psi'(x) = \Lambda_{1/2}\psi(\Lambda^{-1}x)$. As with other transformations, it induces transformations of the creation and annihilation operators, respectively, i.e., it may equivalently be performed by transformations of these operators alone. These transformations, the explicit forms of which are not important here, preserve the anticommutation relations, of course.
Now for my question: Why can a transformation of only the annihilation operators $a^{s}_{\mathbb{p}}$, say, i.e., the positive frequency part of $\psi(x)$, not be performed? Although such a transformation does not 'back-induce' a transformation of $\psi(x)$ itself as any matrix times $\psi(\Lambda^{-1}x)$, as is the case for the 'full' Lorentz transformation, it will still preserve the anticommutation relations for the creation and annihilation operators and thus imply a causal theory.
A related question is the following: Why must, as all text books seem to claim, any symmetry transformation of $\psi(x)$ yield something equal to some matrix times $\psi$, evaluated either at the same spacetime point (for internal transformations like $U(1)$ transformations, and charge conjugation), or at some other spacetime point (for external transformations like Lorentz transformations, parity conjugation, and time reversal)?
